Question Number 114959 by mathdave last updated on 22/Sep/20

long time question proposed by  math abdo  ∫_0 ^∞ ((lnx)/(1+x^2 +x^4 ))dx

Answered by mathdave last updated on 22/Sep/20

solution  let  I=∫_0 ^∞ ((lnx)/(1+x^2 +x^4 ))dx=∫_0 ^1 ((lnx)/(1+x^2 +x^4 ))dx+∫_1 ^∞ ((lnx)/(1+x^2 +x^4 ))dx=A+B  puttting  x=(1/x)  into integral B  I=∫_0 ^1 ((lnx)/(1+x^2 +x^4 ))dx−∫_0 ^1 ((x^2 lnx)/(1+x^2 +x^(4 ) ))dx=A−B  let  A=∫_0 ^1 ((lnx)/(1+x^2 +x^4 ))dx=∫_0 ^1 ((lnx)/(1−x^6 ))dx−∫_0 ^1 ((x^2 lnx)/(1−x^6 ))dx  A=Σ_(n=0) ^∞ ∫_0 ^1 x^(6n) .lnxdx−Σ_(n=0) ^∞ ∫_0 ^1 x^2 .x^(6n) .lnxdx  A=Σ_(n=0) ^∞ (∂/∂a)∣_(a=1) ∫_0 ^1 x^(6n) .x^(a−1) dx−Σ_(n=0) ^∞ (∂/∂a)∣_(a=1) ∫_0 ^1 x^(6n+2) .x^(a−1) dx  A=Σ_(n=0) ^∞ (∂/∂a)∣_(a=1) ((1/(6n+a)))−Σ_(n=0) ^∞ (∂/∂a)∣_(a=1) ((1/(6n+2+a)))  A=−Σ_(n=0) ^∞ (1/((6n+1)^2 ))+Σ_(n=0) ^∞ (1/((6n+3)^2 ))=−(1/(36))Σ_(n=0) ^∞ (1/((n+(1/6))^2 ))+(1/(36))Σ_(n=0) ^∞ (1/((n+(1/2))^2 ))  A=−(1/(36))ψ^1 ((1/6))+(1/(36))ψ^1 ((1/2))...........(1) and   B=∫_0 ^1 ((x^2 lnx)/(1+x^2 +x^4 ))dx=∫_0 ^1 ((x^2 lnx)/(1−x^6 ))dx−∫_0 ^1 ((x^4 lnx)/(1−x^6 ))dx  B=Σ_(n=0) ^∞ ∫_0 ^1 x^2 .lnx.x^(6n) dx−Σ_(n=0) ^∞ ∫_0 ^1 x^4 .lnx.x^(6n) dx  B=Σ_(n=0) ^∞ (∂/∂a)∣_(a=1) ∫_0 ^1 x^(6n+2) .x^(a−1) dx−Σ_(n=0) ^∞ (∂/∂a)∣_(a=1) ∫_0 ^1 x^(6n+4) .x^(a−1) dx  B=Σ_(n=0) ^∞ (∂/∂a)∣_(a=1) ((1/(6n+2+a)))−Σ_(n=0) ^∞ (∂/∂a)∣_(a=1) ((1/(6n+4+a)))  B=−Σ_(n=0) ^∞ (1/((6n+3)^2 ))+Σ_(n=0) ^∞ (1/((6n+5)^2 ))=−(1/(36))Σ_(n=0) ^∞ (1/((n+(1/2))^2 ))+(1/(36))Σ_(n=0) ^∞ (1/((n+(5/6))^2 ))  B=−(1/(36))ψ^1 ((1/2))+(1/(36))ψ^1 ((5/6))........(2)  but  I=A−B  I=−(1/(36))ψ^1 ((1/6))+(1/(36))ψ^1 ((1/2))−(−(1/(36))ψ^1 ((1/2))+(1/(36))ψ^1 ((5/6)))  I=(1/(18))ψ^1 ((1/2))−(1/(36))[ψ^1 ((1/6))+ψ^1 (1−(1/6))]  using  ψ^1 (z)+ψ^1 (1−z)=(π^2 /(sin^2 (πz)))  ψ^1 ((1/6))+ψ^1 (1−(1/6))=(π^2 /([sin((π/6))]^2 ))=4π^2   I=(1/(18))ψ^1 ((1/2))−((4π^2 )/(36))=(1/(18))((π^2 /2))−((4π^2 )/(36))=−((3π^2 )/(36))=−(π^2 /(12))  ∵∫_0 ^∞ ((lnx)/(1+x^2 +x^4 ))dx=−(π^2 /(12))  b8y nathdave(22/09/2020)

Commented byI want to learn more last updated on 23/Sep/20

And sir, does the identity has proof?

Commented byI want to learn more last updated on 23/Sep/20

Weldone sir.  Please help me with this        ∫_( 0) ^( ∞)  ((10 sin(x))/x) dx  Also without the limit.   ∫  ((sin(x))/x) dx.  Thanks sir.

Commented bymathdave last updated on 23/Sep/20

note ∫_0 ^∞ ((sin(ax))/x^n )dx=((πa^(n−1) )/(2Γ(n)sin(((πn)/2))))  ∫_0 ^∞ ((sin(x))/x)dx=((π(1)^(1−1) )/(2Γ(1)sin((π/2))))=(π/2)  where Γ(1)=1,a=1 and n=1  ∵∫_0 ^∞ ((sin(x))/x)dx=(π/2)

Commented bymathdave last updated on 23/Sep/20

u can using IBP for d one without d  symmentric bondaries

Commented byI want to learn more last updated on 23/Sep/20

Thanks sir, i appreciate.

Commented byI want to learn more last updated on 23/Sep/20

I will love to see your work sir

Commented bymathdave last updated on 23/Sep/20

yah it has proof

Commented byI want to learn more last updated on 23/Sep/20

Please sir help me to prove it when you are less busy. Thanks sir. I appreciate.

Commented byTawa11 last updated on 06/Sep/21

great sir

Answered by Bird last updated on 24/Sep/20

let A =∫_0 ^∞  ((lnx)/(x^4  +x^2 +1))dx  we use ∫_0 ^∞  q(x)ln(x)dx  =−(1/2)Re(Σ Res(q(z)ln^2 z ,a_k ))  w(z)=((ln^2 z)/(z^4  +z^2  +1))  pole of w?  u^(2 )  +u +1=0  (u=z^2 )  Δ=−3 ⇒u_1 =((−1+i(√3))/2) =e^(i((2π)/3))  and  u_2 =((−1−i(√3))/2) =e^(−((i2π)/3))   w(z) =((ln^2 z)/((z^2 −e^((i2π)/3) )(z^2 −e^(−((i2π)/3)) )))  =((ln^2 z)/((z−e^((iπ)/3) )(z+e^((iπ)/3) )(z−e^(−((iπ)/3)) )(z+e^(−((iπ)/3)) )))  Res(w,e^((iπ)/3) )=(((((iπ)/3))^2 )/(2e^((iπ)/3) (2isin(((2π)/3)))))  =(1/(4i))×((−π^2 )/9) ×(e^(−((iπ)/3)) /((√3)/2)) =−(π^2 /(18i(√3))).e^(−((iπ)/3))     Res(f,−e^((iπ)/3) ) =(((iπ+((iπ)/3))^2 )/(−2e^(−((iπ)/3)) (2isin(((2π)/3)))))  =((−(((4π)/3))^2 )/(−4i))×(e^((iπ)/3) /((√3)/2)) =((16)/9)π^2 ×(e^((iπ)/3) /(2i(√3)))  Res(f,e^(−((iπ)/3)) ) =(((−((iπ)/3))^2 )/((−2isin(((2π)/3)))2e^(−((iπ)/3)) ))  =−(π^2 /9)×(e^(i(π/3)) /(−4i×((√3)/2))) =(π^2 /(18i(√3))).e^((iπ)/3)   Res(f,−e^(−((iπ)/3)) )  =(((iπ −((iπ)/3))^2 )/((−2e^(−((iπ)/3)) )(−2i sin(((2π)/3))))  =((−(((2π)/3))^2 )/(4i ×((√3)/2))).e^((iπ)/3)  =−((4π^2 )/9)×(e^((iπ)/3) /(2i(√3)))  ⇒Σ Res(f)=−(π^2 /(18i(√3)))e^(−((iπ)/3))   +((16)/9)π^2  (e^((iπ)/3) /(2i(√3)))  +(π^2 /(18i(√3))) e^((iπ)/3)   −((4π^2 )/(18i(√3)))×e^((iπ)/3)   =((iπ^2 )/(18(√3))){(1/2)−((i(√3))/2)}  −((16iπ^2 )/(18(√3))){(1/2)+((i(√3))/2)}−((iπ^2 )/(18(√3))){(1/2)+i((√3)/2)}  +((2π^2 i)/(9(√3))){(1/2)+i((√3)/2)}...be continued...